A331548 15-adic integer x = ...2AA66B44A40E43797853AD13 satisfying x^5 = x; also x^3 = -x; (x^2)^3 = x^2 = A331550; (x^4)^2 = x^4 = A331549.

3, 1, 13, 10, 3, 5, 8, 7, 9, 7, 3, 4, 14, 0, 4, 10, 4, 4, 11, 6, 6, 10, 10, 2, 8, 1, 9, 9, 0, 4, 8, 3, 10, 11, 5, 9, 11, 0, 8, 0, 10, 9, 2, 6, 0, 8, 11, 5, 8, 5, 7, 1, 6, 10, 5, 12, 14, 0, 0, 6, 10, 6, 12, 8, 2, 12, 4, 6, 1, 6, 14, 6, 7, 8, 13, 5, 5, 3, 4, 3, 0

Offset

0,1

Comments

The base-15 version of A120817. A, B, C, D, and E are the standard notations for the hexadecimal digits 10, 11, 12, 13, and 14, respectively.

Conjecture: If k is the number of prime factors congruent to 1 (mod 4) of an integer n, then there are exactly k n-adic integers x satisfying x^5 = x, while not satisfying x^h = x for h = 2, 3, or 4. This does not count -x, which also satisfies, in each case. - Patrick A. Thomas, Mar 31 2020

Links

Table of n, a(n) for n=0..80.

Formula

x = 15-adic lim_{n->infinity} 3^(5^n).

Example

x equals the limit of the (n+1) trailing digits of 3^(5^n):
3^(5^0) = (3), 3^(5^1) = 1(13), 3^(5^2) = 1708EB01(D13), ...
x   = ...2AA66B44A40E43797853AD13.
x^2 = ...65762C0520697E8CA1A31469 = A331550.
x^3 = ...C44883AA4AE0AB75769B41DC = -x.
x^4 = ...8978C2E9CE8570624D4BDA86 = A331549.
x^5 = ...2AA66B44A40E43797853AD13 = x.

Prog

(PARI) \\ after Paul D. Hanna's program in A120817
{a(n)=local(b=3, v=[]); for(k=1, n+1, b=b^5%15^k; v=concat(v, (15*b\15^k))); v[n+1]}
for(k=0, 80, print1(a(k), ", ")) \\ Hugo Pfoertner, Jan 26 2020
(PARI) (A331548_vec(n)=Vecrev(digits(lift(Mod(3, 15^n)^5^(n-1)), 15)))(99) \\ M. F. Hasler, Jan 26 2020

Crossrefs

Cf. A120817, A331549, A331550.

Sequence in context: A088729 A270968 A142888 * A068439 A133171 A133177

Adjacent sequences: A331545 A331546 A331547 * A331549 A331550 A331551

Keyword

base,nonn

Author

Patrick A. Thomas, Jan 20 2020

Status

approved